Consider a fixed charge Q1. Let us move a second charge say test charge Qt around the Q1. We observe that the test charge Qt experiences force everywhere around Q1. It is experiencing a force field. Force on it will be given by coulombs law as:

But still question arises what is Electric Field?? Before defining electric field let us modify above equation a bit.
Let us write the above equation as force/charge equation:

This quantity that is force/charge describes a vector quantity. This vector quantity is called Electric Field Intensity.So Electric field intesity is defined as the vector force on an unit positive test charge by a fixed charge Q1. Or it can also be defined as force per unit charge.
So to sum it up we can Electric field intensity as:

So it becomes.

Expressing Electric field in cartesian Coordinate:
![E(r) = \frac{Q}{4\pi \epsilon _{0}\left | r - r^{'} \right |^{2}}\frac{r - r^{'}}{\left | r - r^{'} \right |} = \frac{Q[(x - x^{'})i + (y - y^{'})j + (z - z^{'})k]}{4\pi \epsilon _{0}[(x - x^{'})^{2} + (y - y^{'})^{2} + (z - z^{'})^{2}]^{\frac{3}{2}}}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ux0ib0FdFLzIiBT1AfOZciq-L7F1fYJR7TDBgf6n4spHJSG0sph8JhTnQ-uzgyNAEeqE_FoNGLN2_uNOn6S8PfoQ_VBVRXhfnNSextBaKfd9D7xZXhbuNTXhzY89vbYpTpGGYB-gFd3CGMgLHLyAEQINCMjamuWbdpKfRoBjEKG24p76cptH4joYIFl7a_FYxB_45fVlw1ZbT2y86VPlG0MgdSEx34IPcp0_FtGUrXr1jAzq0VNuk8vSnPZFHunmHWEEZlXJyUalIq2PvQwaagoZMA0MAP6hZCJzbEb5U24VOBSvW-mwLAAB98OnJfVqJhZyq8FP-tdgQo1RGLP3JRvnd2dUHbIfP2WnI6QWyjOcrrFjfB18OsBjyPsmBc6B71o0mioHlhcGw7lwbzgxismNholbo8468y6duQEPIX0K7U-Ugb-wBHPoyDzRoMK8e2tb9ddvlOlf2yGcwZa2ErOnXww9jS6AwbUOwOTiJfXpPLTyt1WbcBrUj-wfLCyahYWKAGdQ5mqbmT2Kz9EUP3PhWVYzzBE9AufAsC3607vAze0ETH6zyTqIWNC6Ax8f8vmYmuiReXguQQGGsf6i9aiG-CryXoemsnx6AqN07cXa-qa0M9gQCk8HDkZjTaCPmksCq0A467QeA=s0-d)
Expressing Electric Field in Spderical Co-ordinates:
First for considering spherical co ordinate we must see that fixed charge has spherical symmetry at its location like at origin(0,0,0) for that we can either use spherical co-ordinate directly in equation as:

Where Er and r spherical co-ordinates values or we can convert them in cartesian coordinate which will give us the equation as below:
We will have r = R = xi + yj +zk ar = aR = (xi + yj + zk)/√(x^2+y^2+z^2 )

If more charges are added in the surrounding of Q1 then electric field will be sum of individual acting alone. As shown in equation given below:

Before ending the topic let us do a small quick example.
Example: See the figure below. We need to find out Electric field at P(1,1,1) caused by 4 identical 3nC charges located at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0).
[caption id="" align="alignnone" width="615" caption="Example of Electric Field"][/caption]
Answer: As r = i + j + k and r1 = i+j therefore r – r1 = k. Therefore |r – r1| = 1.
Similaraly r – r2 = 2i + k. therefore |r – r2| = √5
Similaraly r – r3 = 2i + 2j + k. therefore |r – r3| = 3
Similaraly r – r4 = 2j + k. therefore |r – r4| = √5
And as

![E = 26.96[\frac{k*1}{1*1^{2}}+\frac{2i + k}{\sqrt{5}}\frac{1}{\sqrt{5}^{2}} + \frac{2i + 2j + k}{3}\frac{1}{3^{2}}+ \frac{2j + k}{\sqrt{5}}\frac{1}{\sqrt{5}^{2}}]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sj1ZoGPTs6GALpXeymTrHpa0wLfo8TGCcSU00EOflGbltz_gmfgm-w2v9oMyYXNwz1i1N2UUj2qkswx6WTteA1qBejpyaTaQuOUHY168-3sIjqpZY6GUWPyb_V8CMlYmIOv0TOWIHZkbrbhn9w45dEOom4fP8dYh_3jpJ10ZWRLajDO1c7Kt7_euYFuObX_yUXEcBFXCzd0HLiNkpxLedP3hk3ODzEWmGFoKr7OXYI7lcotGnVE90nqv5TC8xzzuPZ-8TShXuucI1n1axBf6pF5fsI_Fa1C3rgR5oL7vyhRnK3dY0PeWC8pfn29HtsyhRGY9HoVYbwEih7oRiLQjUf6X0SnCfEuJ8LZlPLTkoCuqeSft0NEobAjVn56ChFZox0tdYVhvoCOGfOipgiNkUnT9rMS7u20-mdEU7nJhthwSXLHdHlF3aybbCJEHMc1LC86yzpxaEwTQ_8V42FCkFENJaa0w=s0-d)
Therefore simplifying we get E =
E = 6.82i + 6.83j + 32.8k V/m ans.
But still question arises what is Electric Field?? Before defining electric field let us modify above equation a bit.
Let us write the above equation as force/charge equation:
This quantity that is force/charge describes a vector quantity. This vector quantity is called Electric Field Intensity.So Electric field intesity is defined as the vector force on an unit positive test charge by a fixed charge Q1. Or it can also be defined as force per unit charge.
So to sum it up we can Electric field intensity as:
So it becomes.
Expressing Electric field in cartesian Coordinate:
Expressing Electric Field in Spderical Co-ordinates:
First for considering spherical co ordinate we must see that fixed charge has spherical symmetry at its location like at origin(0,0,0) for that we can either use spherical co-ordinate directly in equation as:
Where Er and r spherical co-ordinates values or we can convert them in cartesian coordinate which will give us the equation as below:
We will have r = R = xi + yj +zk ar = aR = (xi + yj + zk)/√(x^2+y^2+z^2 )
If more charges are added in the surrounding of Q1 then electric field will be sum of individual acting alone. As shown in equation given below:
Before ending the topic let us do a small quick example.
Example: See the figure below. We need to find out Electric field at P(1,1,1) caused by 4 identical 3nC charges located at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0).
Example of Electric field |
Answer: As r = i + j + k and r1 = i+j therefore r – r1 = k. Therefore |r – r1| = 1.
Similaraly r – r2 = 2i + k. therefore |r – r2| = √5
Similaraly r – r3 = 2i + 2j + k. therefore |r – r3| = 3
Similaraly r – r4 = 2j + k. therefore |r – r4| = √5
And as
Therefore simplifying we get E =
E = 6.82i + 6.83j + 32.8k V/m ans.
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